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Seemingly easy physics problem

  1. Jun 4, 2004 #1
    A door, 2.30m high and 1.30m wide, has a mass of 13.0kg. A hinge 0.40m from the top and another hinge 0.40m from the bottom support half the doors weight. Assume the CM is at the geometrical centre of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

    The trick with the horizontal forces...

    think of breaking the bottom hinge, and there is no door frame. Which way would the bottom of the door swing? to the right, if you use the picture. If you break the top hinge but leave the bottom one in tact, then the top of the door would swing to the left. This means the x-force for the bottom points to the left and the x-force for the top points to the right...and they must be equal since the door doesn't swing at all. Use net force to analyze that.

    How can i go from here?

    This problem is giving me a hard time for some reason.
     
  2. jcsd
  3. Jun 4, 2004 #2

    jamesrc

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    Gold Member

    You need to use the fact that the door is in equilibrium to write equations that will allow you to find the forces in question. Here Are the labels I will use:

    W = mg = weight of door
    Fxu = horizontal reaction force exerted by upper hinge on door
    Fxl = horizontal reaction force exerted by lower hinge on door
    Fyu = vertical reaction force exerted by upper hinge on door
    Fyl = vertical reaction force exerted by lower hinge on door
    d = distance between the two hinges = 2.3 - .4 - .4
    r = half the width of the door = 1.3/2

    First consider all of the forces in the y-direction:

    You are given that Fyu = Fyl = W/2, which fits with the sum of forces in the y-direction equalling 0:

    [tex] \Sigma F_y = 0 = F_{yu} + F_{yl} - W [/tex]

    Now the forces in the x direction:

    [tex] \Sigma F_x = 0 = F_{xu} + F_{xl} [/tex]

    (these two forces are assumed to be in the same direction; obviously they are actually equal in magnitude and opposite in direction which is borne out by the math)

    Now the moments about a selected point; let's pick the bottom hinge:

    [tex] \Sigma \tau_{l} = 0 = -F_{xu}d + Wr [/tex]

    (I'm using a right-handed coordinate system and in my picture, the hinges are on the right-hand side of the door, as implied by your description of breaking hinges. Be sure to work everything out for yourself and keep your signs straight.)

    Hopefully, I didn't make any mistakes and you can follow that setup to the answer. Let me know if anything is unclear.
     
  4. Jun 4, 2004 #3
    Ack that torque equation was surprisingly tricky but it all makes sense. Thank you so much james.
     
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